Prove ( a + b + c )^2 <= 4( ab + bc + ca )

[Bernie Hunt]

The Problem;
If a>0, b>0, c>0 and a + b > c, b + c > a, c + a > b then ( a + b + c )^2 <= 4( ab + bc + ca ).

Hints: Start with ( a + b + c )^2 and establish the inequality ( a + b + c )^2 <= 4( ab + bc + ca ). Us the inequality fact: if x < y then xy < y^2 if y > 0.

I haven’t make much progress. I’ve been trying;

( a + b + c )^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca

But I haven’t been able to figure out a way to get the right side manipulated. I tried a couple of substitutions, but couldn’t justify a<b or such to be able to use the hint.

Can anyone give me a push in the right direction?

Thanks,
Bernie

 

[acm]

I would start the question by expanding the RHS of the equation to:
a^2 + b^2 + c^2 + 2(ab + bc + ac)

Now, It is obvious that (a + b + c)^2 >= 2(ab + bc + ac)
Hence 2(a + b + c)^2 >= 4(ab + bc + ac)
Therefore we can assume that 4(ab + bc + ac) = 2(a + b + c)^2
And as (a + b + c)^2 <= 2(a + b + c)^2
Therefore (a + b + c)^2 <= 4(ab + bc + ac)

 

[Bernie Hunt]

ACM,

Thanks for the quick reply. I think there is a flaw in your logic. The statements workout to;

2(a + b + c)^2 > (a + b + c)^2 > 2(ab + bc + ac)
2(a + b + c)^2 > 4(ab + bc + ac) > 2(ab + bc + ac)

Therefore 4(ab + bc + ac) > (a + b + c)^2

Which is saying that:
x > y
x > z
Therefore y > z

Which is not always true. It is true sometimes but not for all conditions.

Is there a way to prove that because a + b > c, b + c > a, c + a > b then the above logic is always true?

Bernie

 

[StatusX]

Expand out the square and cancel like terms. Then use their clue.

 

[acm]

No problem, Should have though that through.

 

[matt grime]

The Problem;
If a>0, b>0, c>0 and a + b > c, b + c > a, c + a > b then ( a + b + c )^2 <= 4( ab + bc + ca ).

Hints: Start with ( a + b + c )^2 and establish the inequality ( a + b + c )^2 <= 4( ab + bc + ca ). Us the inequality fact: if x < y then xy < y^2 if y > 0.

I haven’t make much progress. I’ve been trying;

( a + b + c )^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca

But I haven’t been able to figure out a way to get the right side manipulated. I tried a couple of substitutions, but couldn’t justify a<b or such to be able to use the hint.

Can anyone give me a push in the right direction?

Thanks,
Bernie

Either a=b and you can use that fact, or either a<b, or b<a and wlog we may assume that a<b in this case. (or, they're either all equal, when it is trivially true, or we may assume a<b without loss of generality - the labels of a,b,c are obviously not important).

 

[Bernie Hunt]

StatusX,

Expand out the square and cancel like terms. Then use their clue.

Cancel out which like terms? Expanding out the square gives;
( a + b + c )^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca

Which of these terms can be canceled?

Bernie

 

[StatusX]

With the other side.

 

[durt]

And then rewrite 2ab as ab+ab, etc. Also, I don't think equality ever holds, unless you meant a + b >= c, b + c >= a, c + a >= b.

 

[Bernie Hunt]

OK, here we go;

Prove:
If a>0, b>0, c>0 and a + b > c, b + c > a, c + a > b then ( a + b + c )^2 <= 4( ab + bc + ca ).

Hints: Start with ( a + b + c )^2 and establish the inequality ( a + b + c )^2 <= 4( ab + bc + ca ). Us the inequality fact: if x < y then xy < y^2 if y > 0.

(a+b+c)^2 = (a+b+c)^2 (Reflective property of equality)

(a+b+c)^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca

a + b > c
c(a+b) > c^2 fact of inequity
ca + cb > c^2

b + c > a
a(b+c) > a^2 fact of inequity
ab + ac > a^2

c + a > b
b(c+a) > b^2 fact of inequity
bc + ab > b^2

From above
(a+b+c)^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca

By substitution
(a+b+c)^2 < (ab + ac) + (ca + cb) + (bc + ab) + 2ab + 2bc + 2ca

(a+b+c)^2 < 2ab + 2ac + 2bc + 2ab + 2bc + 2ca
(a+b+c)^2 < 4ab + 4ac + 4bc
(a+b+c)^2 < 4(ab + ac + bc)

(a+b+c)^2 <= 4(ab + ac + bc) ?

Can I make this last step from < to <=? If so is there a principal or property that justifies it?

Thanks,
Bernie

 

[radou]

Think about what happens if a = b = c = 0.

 

[StatusX]

If something is less than something else, it is certainly also less than or equal to it. It's probably a typo though, and you were supposed to prove your next to last line.

 

[Bernie Hunt]

The proof is definitely <=, but I think I found the justification. I don't have my notes handy, but I think it's disjunctive addtion justifies that if p then (p or q). Also if (p or q) and (~q) then p.

Bernie

 

[StatusX]

Yea, although it's pretty obvious if you just think about it. Also, I'm not saying you copied the question wrong, I'm saying it was wrong in the book. It's not strictly incorrect, but it's completely unnatural to use a $\leq$ in place of a < when equality is not possible.